Sistem Bilangan
Oleh :
Moch Nur Purnama
Analogue vs Digital
Analogue
* Continuous range of value
* Precision limited by Noise Digital
* Discrete range of values
* Precision limited by number of “Bit”
Analogue vs Digital
Analogue Digital
Analogue vs Digital
The real world is analogue ( by because all signal in world be shape analogue)
But in controlling, Digital one had using for process.
Both of signal had been converter each other
Analoge vs Digital
Analogue A to DDigital
ProcessingD to A Analogue
Why Digital Only by using in Processing?
^ Adventure in integrated Circuit has made the complex processing of digital data. ^ Digital Control processing has made easier than analogue^ Digital circuits are inherently more noise resistant
Digital and Boolean
Digital represented by boolean logicBoolean is the name of mathematician’s
expert Now boolean is called by conventional
logic because there is new logic that called by fuzzy logic
But all electronic still using boolean logic to processing the controlling system
Why Boolean
It is convenient in electrical system to use a two-value system to represent value true/false, on/off, yes/no and 1/0* Two voltage or current levels can be used* Easier to process and distribute reliably (diandalakan)* Don’t think of them as numbers (even though we often represent them as 0/1 for brevity(ketangkasan))
The need for binary numbers* Multi-value quantities need to be represented in the digital system. Therefore need numbers made up from the simple two value system
Positional Number System
3578.778
8x10-3
8 x 100
7 x 101
5 x 102
3 x 103
7x10-2
7x10-1Decimal point
Base 10, weigthing are powers of 10
Unsigned binary numbers
1100.101
1 x 2-3 = 0.125
0 x 20= 0.0000 x 21= 0.0001 x 22= 4.0001 x 23= 8.000
0 x 2-2 = 0.000
1 x 2-1 = 0.500Binary point
Each bit of the Number may beRepresentaed byA Boolean value
Binary, weightings are powers of 2
Multi-precision Arithmatic
Additional of A and B
A1 B1
A2 B2
+
B3
Carry Flag
Carry Flag
A2
Carry Out
Carry In
Carry Out
Multi-precision Arithmatic
A1 B1
A2 B2
-
B3
Carry Flag
A2
Carry Out
Carry In
Hexadecimal Numbers
660: 164
41: 169
2
Hexadecimal : 294 Hex
0123456789101112131415
0123456789ABCDEF
215: 1613
7
Hexadecimal : 7D Hex
Hexadecimal Numbers
660 0010 1001 01000123456789ABCDEF
0000000100100011010001010110011110001001101010111100110111101111
2 9 4
215 0000 1101 0111
0 D 7
Decimal to Binary
Number = 36.375
Base = 2
Decimal Number
Binary Digits
Converter Number
0 0 0100100.0110
0.5 1 0100100.011
0.75 1 0100100.01
0.375 0 0100100.0
36 0 0100100
18 0 010010
9 1 01001
4 0 0100
2 0 010
1 1 01
0 0 0
Generetee each digit by successive division Or multiplication.
There is no guarantee the fraction will be finite
Fractional part – Multiplication by base
Whole part – divition by base
Binary Additional
0 + 0 = 0
0 + 1 = 1
1 + 0 = 1
1 + 1 = 0 carry 1
Easy Layaou ?
Binary Addition
190 + 141 =331
1 0 1 1 1 1 1 01 0 0 0 1 1 0 1
110
1
1
1
0
1
0
1
101
Carry out of8-bit number
1
Carry out ofEach column
Binary Subtraction
229 – 46 = 183
1 1 1 0 0 1 0 1
0 0 1 0 1 1 1 0
1
1
2
1
1
2
10
1
2
1
2
1
1
2
101 Borrow out
Borrow in fromLeft column
A borrow-out of 1 fromThis column becomes a borrow in of 2 in this column
Both rows subtracted
Exercise
Convert to 8-bit binary and do the arithmetic operation
* 120 + 54 * 110 + 100
* 224 – 134 * 200 + 20
* 112 – 89 * 111 – 25 Convert back to decimal and check the
result
Binary Number Circle
4 – bitBinary
Number Circle
In real hardware there is a fixed numberOf bits available. We often ignore leading zerosBut they are still there!
Examlpe :If we only use 4 bits then the binaryCounting sequence “wraps around”At 15 ↔ 0
11 - 1 = 10
11 1110- 1 1
10 1010
Binary Number Circle
4 – bitBinary
Number Circle
Subtracting across the boundaryStill “works” if you think of result As the distance on the number Circle.
(Module arithmetic – ignoreThe borrow /carry)
8 1000- 14 - 1110
10 (-1)1010
Representing –ve Number
Several choices for natation
* sign + magnitude notation
* 1’s complement
* 2’s complement notation
* various ‘excess codes ‘
Sign Number – sign + magnitude Notation
Sign Bit Magnitude
0 +ve Simple binary number
1 - ve
Problem ?
How about Null or Zero
+ 0 0000 - 0 1000
Signed Numbers – Sign + magnitude Notation
Arithmetic Difficult to do – have to work out that operation to
perform 5 + -6 actually calculate –(6-5) i.e. exchange the
operands and do subtraction! -5+ -6 actually calculate –(5+6) i.e. negate the addition
of the negated numbers ! Required action depends the signs of the numbers
and on which has the large magnitude. Natural for us –a bit hard for the computer since the only way it can work out the bigger number is to do a subtraction!
Sign + Magnitude Examples
Value4-bit sign + magnitude
8-bit sign + magnitude
+7 0111 00000111
+6 0110 00000110
…… …… ……
+1 0001 00000001
+0 0000 00000000
-0 1000 10000000
-1 1001 10000001
-2 1010 10000010
…… …… ……
-7 1111 10000111
Sign Numbers – 2’s Complement
As for straight binary numbers but with the weighting of the most significant bit being negative
Example
* 4 bit – weights are -8, 4,2,1
* 8 bit – weights are -128, 64,32,16,8,4,2,1 Need to know how many bits are being used
to work out the value of the number – don’t omit leading zeroes
Sign Numbers – 2’s Complement
1100.101
1 x 2-3 = 0.125
0 x 20= 0.0000 x 21= 0.0001 x 22= 4.0001 x 23= -8.000
0 x 2-2 = 0.000
1 x 2-1 = 0.500Binary point
Sign Bit
Binary, weightings are powers of 2 -4.375
2’s Complement Examples
Value4-bit sign 2’s complement
8-bit sign complement
+7 0111 00000111
+6 0110 00000110
…… …… ……
+1 0001 00000001
+0 0000 00000000
-1 1111 11111111
-2 1110 11111110
…… …… ……
-7 1001 11111001
-8 1000 11111000
2’s Complement Examples
Example : -4 (decimal)
Become 4 = 0100 ( binary) = 1x22 = 42’s Complement -4= 1100 (binary) = -(23) + 22
= -8 + 4 = -4
Exercise
Converse decimal number above into negative (2’s complement) :
1. -7 ( 4 digit ) 6. 6 (4 digit)
2. -7 (8 digit) 7. 10 (8 digit)
3. -12 (8 digit) 8. 30 (8 digit)
4. -20 (8 digit) 9. 98 (digit)
5. -100 (8 digit) 10. 126 (digit)
Addition 2’s Complement
For 4 digit :
4 0100
3 + 0011 +
7 0111
22+21+20 = 4+2+1 =7
Addition 2’s Complement
For 4 digit
-1 1111
-2 + 1110 +
-3 11101
-(8)+4 +0 + 1 = -3 Carry out
Exercise
For 4 Digit : 1. 7 + (-5)2. -6 + -13. 3 + 44. 2 + 35. -4 + 7Converse all item to digital and addition.
And then Converse to decimal again
Subtraction 2’s Complement
+ 7 0111
+ 3 (0011)- 1101 +
+4 10100
Discard
Subtraction 2’s Complement
(-8) 1000
(-3) = 1101 - 0011 +
-5 1011
Exercise
for 4 digit . Converse decimal above to digit and subtraction. After that converse to decimal again :
1. (+3) – (-3)2. (-4) – (+2)3. (-8)- (+4)4. (-3) – (-4)5. (7) – (5)
2’s Complement ALU
Addition and subtraction use the same rules as unsigned binary.
Same hardware may be used for both Carry (C) is used for unsigned, overflow (v) for signed
C=Carry
V=overflow
OP
Signed Numbers
Signed Numbers
C=Carry
V=overflow
OP
Signed Numbers
Signed Numbers
The samehardware
Arithmetic Flags in Condition code register (CCR)